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The no. of ways in which 6 persons can be seated in a circular table so that two particular persons are never together is?

$\begin{array}{1 1} 72 \\ 120 \\ 240 \\ 480 \end{array}$

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  • No. of arrangement of n different things in a circle =$(n-1)!$
Total no. of ways without any condition =$5!$
No. of ways in which two particular persons are together=$4!.2!$
(i.e., assume the two persons to be 1 then there will be 5 persons (4! ways) and
the two persons can interchange their positions in 2! ways.)
$\therefore $ no. of ways in which two persons are never together is
$5!-4!.2!=120-48=72$

 

answered Aug 13, 2013 by rvidyagovindarajan_1
edited Dec 20, 2013 by meenakshi.p
 

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